1 Name: _____________________________________________________________________________ Genetics Unit 7; Chapters 10 & 11 Date Classwork Homework PTC Tasting Begin Notes (PowerPoint) Reading and Questions for 10.2 Notes Reading and Questions for 10.3 Finish Notes Start Monohybrid Crosses Wkst (pages 14-17) Monohybrid Cross Worksheet Go over Monohybrid Crosses Review Notes Start Dihybrid Crosses Wkst (pages 18-19) Review Monohybrid crosses Dihybrid crosses Review Dihybrid Crosses “Ghost in Your Genes” Video Reading and Questions for 11.1 More Punnett Square Practice Wkst (pg 20-21) Finish “Ghost in Your Genes” Go over “More Punnett Square Practice” Wkst Reading and Questions for 11.2 Faces Lab Reading and Questions for 11.3 Work on Faces Lab Packet Check Day Finish any part to the lab Finish Faces Pictures and Present Read Karyotyping Lab Magnetic Karyotyping lab Additional Epistasis Problem (pg 28) Study for Test/ Prepare Packet Corn Activity (if time) Go over Epistasis Problem Review Day Review for test Test and Packet Due Rubric: Out of Your Score Genetics Reading Questions 20 Notes 40 Monohybrid Crosses Wkst 20 Dihybrid Crosses Wkst 20 More Punnett Square Practice Wkst 20 Faces Lab (chart) 20 Epistasis Problem Wkst 10 Karyotyping Lab 30 Corn activity 20 Total
Heterozygous- two different alleles Genotype- an organisms
________________ of alleles Phenotype- ____________________
characteristic
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Mendel’s Law of Segregation
Two alleles ____________________ during meiosis You have a ____________________ chance of
giving each allele
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Monohybrid Cross
Cross a homozygous dominant plant and with a homozygous recessive plant tall - dominant dwarf - recessive
What do we expect the F1 and F2 Generations to look like?
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Dihybrid Cross
Simultaneous inheritance of 2 traits in the same organism
Must be on __________________ chromosomes
Different chromosomes separate __________________
Genes on different chromosomes separate independently
Dihybrid Cross Law of Independent Assortment
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Cross two pea plants that are each hybrid for height (Tt) and also hybrid for pod color
•(T-Tall and t- dwarf)•(Y=yellow and y=green)
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What would happen?
What would happen if two genes were on the same chromosome?(Circle one) A. They sort
independently of each other
B. They don’t sort independently of each other
Why?
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Complex Inheritance and Human Heredity
Albinism Can affect only the eyes, skin, or hair or be
complete
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Sickle cell
Complex Inheritance and Human Heredity
Sickle Cell Anemia
__________________ Recessive
Red blood cells- sickle shape heterozygotes have both
normal and sickle-shaped cells Heterozygotes protected from
malaria HH- no SCA/ get malaria Hh- no SCA/ don’t get malaria hh- afflicted with SCA
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Pedigrees
A diagram that traces inheritance Can be used to predict __________________
__________________ in families
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Incomplete Dominance
The heterozygote phenotype is a __________________ between the two homozygous phenotypes
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Codominance
__________________ alleles are expressed at the same time
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No dark pigment present in fur Dark pigment present in fur
eebbeeB_ E_bb E_B_
Complex Inheritance and Human HeredityEpistasis
One ______________ can ______________ the effects of another allele
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Sex Determination (mammals)
Sex Chromosomes XX- ______________ XY- ______________
Sex chromosomes determine ______________
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Sex Linked Traits
Located on the _______ chromosome
Affects mostly ______________
Examples: Red Green
Colorblindness Hemophilia
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Polygenic Traits
Trait determined by ______________ genes at many loci ______________– locations on different chromosomes
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Environmental Influences
Diet Exercise Sunlight and water Temperature
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Identical Twin Studies
Helps separate ______________ contributions from ______________ contributions
Traits that affect both twins are controlled partially by heredity
Traits expressed differently in identical twins are strongly influenced by the ______________
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Karyotype Studies
Picture of the chromosomes arranged in decreasing size
Chromosomes are ______________ Look for abnormalities
Number of chromosomes Missing parts to chromosomes
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Nondisjunction
During meiosis- sister chromatids ______________ to separate ______________ Down syndrome Sex Chromosomes
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Instructions Genetic Problem Format
Use this format for your genetics problems:
1. Make a Key a. Use a capital letter for the dominant allele b. Use a lower case letter for the recessive allele c. Example: S = six fingers; s = five fingers
2. Write the parental cross
a. Example: SS x Ss
3. Do the Punnett Square
4. Give the phenotypic and genotypic ratios a. Phenotype: _______ % dominant (don’t write “dominant”, list the trait)
_______ % recessive (don’t write “recessive”, list the trait) b. Genotype: _______ % pure dominant/homozygous dominant
_______ % hybrid dominant/heterozygous dominant _______ % pure recessive/homozygous recessive
5. Answer the question asked in the problem
Remember:
• Homozygous = Pure = AA or aa • Heterozygous = Hybrid = Aa
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Monohybrid Crosses
1. Albinism is the absence of skin pigmentation and is a recessive trait found in humans and other animals. In the human population about 1/20,000 individuals is an albino. Normal pigmentation (A) is dominant to albinism (a). If an albino woman marries a homozygous normal man, what is the likelihood that one of their children will display albinism?
KEY: __________________________________________________________________ PARENTAL GEOTYPES: ♀ ________________ ♂ ________________ POSSIBLE GAMETES (eggs and sperm): ♀____ and _____ ♂ _____ and _____ SET UP AND FILL IN THE PUNNETT SQUARE: ♂ ♀
PHENOTYPES: ____________________________________________________________________ GENOTYPES: ______________________________________________________________________ ANSWER THE QUESTION: ___________________________________________________________ BONUS: A woman with normal pigmentation marries an albino man and their first child is an albino. What are the genotypes of the couple?
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2. A common squash in Texas is the Yellow crooked-neck squash. This fruit is a source of vitamin A, B, and C. It also contains calcium and iron. Yellow colored squash is recessive to white-colored squash. If a yellow male squash is crossed with a heterozygous female white-squash, what are the predicted genotypes and phenotypes of offspring?
KEY: __________________________________________________________ PARENTAL GEOTYPES: ♀ ________________ ♂ ________________ POSSIBLE GAMETES (eggs and sperm): ♀____ and _____ ♂ _____ and _____ SET UP AND FILL IN THE PUNNETT SQUARE: ♂ ♀
PHENOTYPES: ____________________________________________________________________ GENOTYPES: ______________________________________________________________________ ANSWER THE QUESTION: ___________________________________________________________
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3. Polydactyly is a human trait where humans have six fingers on each hand. Polydactyly is dominant over the recessive trait for five fingers. A man is heterozygous for polydactyly marries a normal 5 fingered woman.
a. How many fingers does the man have? _____________________
b. What is the chance that their first child will have 5 fingers?
4. A cross between two pea plants produces 120 offspring. Both are heterozygous for pod color. Green is dominant over yellow. How many of the offspring are expected to be homozygous recessive or have yellow pods.
5. In cavies (guinea pigs) black coat is dominant over white coat (not albinism). Two heterozygous cavies are mated.
a. What are the phenotypes?
b. What are the genotypes?
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6. A cavy breeder has a black cavy. How could the breeder figure out if the cavy is homozygous or heterozygous for coat color?
7. In garden peas the allele for tall plants (T) is dominant to the allele for dwarf plants (t). You cross one heterozygous plant with a homozygous recessive plant. If a total of 148 plants result, how many can be expected to be tall plants, and how many can be expected to be dwarf plants?
8. In fruit flies Wild type (W) wings are dominant over vestigial wings (w). You cross homozygous dominant flies with homozygous recessive flies. What is the genotype and phenotype ratio of your F2 generation. (hint you need 2 Punnett squares).
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Dihybrid Crosses
1. About 70% of Americans perceive a bitter taste from the chemical phenylthiocarbamide (PTC). The ability to taste this chemical results from a dominant allele (T) and not being able to taste PTC is the result of having two recessive alleles (t). Albinism is also a single locus trait with normal pigment being dominant (A) and the lack of pigment being recessive (a). A heterozygous pigmented woman who cannot taste PTC marries a homozygous, normally pigmented man who is a heterozygous taster. What are the genotypes of the possible children?
2. Wolves are sometimes observed to have black coats and blue eyes. Assume that these traits are controlled by single locus genes and are located on different chromosomes. Assume further that normal coat color (N) is dominant to black (n) and brown eyes (B) are dominant to blue (b). Suppose the alpha male and alpha female of a pack (these are the dominant individuals who do most of the breeding) are black with blue eyes and normal colored with brown eyes, respectively. The female is also heterozygous for both traits. What are the possible outcomes for their offspring?
3. Suppose in a strain of soybeans, high oil (H) content in the seeds is dominant to low oil content and four seeds (E) in a pod is dominant to two seeds in a pod. A farmer crosses two soybean plants, both with high oil content and four seeds per pod. The resulting F1 offspring have a phenotypic ratio of 9:3:3:1 (High oil / four seeds : High oil / two seeds : Low oil / four seeds : Low oil / two seeds). What genotype were the parent plants?
4. Directions: For each of the described situations, complete the following: a. List the genotype of each parent described b. Use a Punnett square to predict the possible outcomes of a cross between the 2 parents. c. List the genotypes and the number of that genotype present in the offspring for each cross (can write as
a ratio). d. List the phenotypes and the number of that phenotype present in the offspring for each cross (can write
as a ratio).
Symbols (for pea plant traits): T: tall R: round seeds Y: yellow seeds P: purple flower t: short r: wrinkled seeds y: green seeds p: white flower
1.) Cross a heterozygous tall and round seed pea plant with another heterozygous tall, round seed plant.
2.) Cross a pea plant that is heterozygous for purple flowers and homozygous dominant for yellow seeds with a plant that is heterozygous for purple flowers and homozygous recessive for green seeds.
3.) Cross a plant that is heterozygous for round seeds and homozygous recessive for white flowers with a plant that is homozygous recessive for wrinkled seeds and heterozygous for purple flowers.
4.) Cross a plant that is heterozygous tall and homozygous for green seeds with a plant that is short and is also homozygous for green seeds.
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More Punnett Square Practice For each problem, include a key, the parental cross, the Punnett Square, genotypic ratios and phenotypic ratios. Monohybrid Crosses
1. Freckles are a dominant trait. Patty is homozygous dominant for freckles, while Charlie is homozygous for no freckles. Predicting the probability if their children will have freckles.
2. Larry and Lola Little have achondroplasia, a form of dwarfism that is dominant trait. Both are heterozygotes. Their son, Big Bob Little, is 7’1”. Use a Punnett square to show how Big Bob got his genotype.
Sex-Linked Crosses
1. Hemophilia is a rare heredity human disease of the blood. The blood of individuals with this condition does not clot properly. Without the capacity for blood clotting, even a small cut can be lethal. In a marriage of two nonhemophiliac parents, a bleeder son is born. What are the probabilities of these parents giving birth to sons being bleeders, and to daughters being bleeders?
2. In humans colorblindness is an example of a sex-linked recessive trait. In this problem, a male with colorblindness marries a female who is not colorblind but carries the allele. Using a Punnett square, determine the genotypic and phenotypic probabilities for their potential offspring.
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Dihybrid Crosses
1. In horses, the coat color black is dominant over chestnut. The trotting gait is dominant over the pacing gait. If a homozygous black pacer is mated to a homozygous chestnut, heterozygous trotter, what will be the ratios for genotype and phenotype of the F1 generation?
2. Imagine that a couple is planning to have children. The male is heterozygous for Huntington’s disease and homozygous dominant for Tay-Sachs. The female is homozygous recessive for Huntington’s disease and heterozygous for Tay-Sachs. The couple is curious about the possibility and probability of their offspring inheriting Tay-Sachs and/or Huntington’s. For humans, Huntington’s disease is dominant (H) over the “normal” condition (h), and the “normal” condition is dominant (T) over Tay-Sachs (t). Complete a Punnett square for this cross and record the probabilities for genotypes and phenotypes of the offspring as ratios.
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“Ghost in Your Genes” by Nova
FIRST, read through all these notes. Then fill in the blanks during the video.
Completed in 2003, the Human Genome Project (HGP) was a 13-year project coordinated by the U.S. Department of Energy and the National Institutes of Health. During the early years of the HGP, the Wellcome Trust (U.K.) became a major partner; additional contributions came from Japan, France, Germany, China, and others. Project goals were to:
• identify all the approximately 20,000-25,000 genes in human DNA, • determine the sequences of the 3 billion chemical base pairs that make up human DNA, • store this information in databases, • improve tools for data analysis, • transfer related technologies to the private sector, and • address the ethical, legal, and social issues (ELSI) that may arise from the project.
Though the HGP is finished, analyses of the data will continue for many years. Fill in:
1. Human Genome Project originally had a human gene count of _____________ genes.
2. Turns out we have only about ______________ genes, like _____________ and ___________ but less than ______________; how come, since we are so complex?
3. We are ________% identical to chimps; maybe genes are not the only thing that makes us …us.
4. Angelman’s Syndrome has these symptoms
5. This is due to a deletion of chromosome #___________
6. BUT the same deletion was associated with Prader-Willi Syndrome: the symptoms of this are:
BUT it is even more interesting: if deletion gene was from father, it was P-W and if from mother it was Angelman’s. DNA sequence was the same but the expression was due to some other influence on the gene.
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7. ___________________ molecules attach onto DNA and regulate them. Other molecular tags attach to the histones and wind them tighter or looser to control DNA expression. These are part of the epigenome. They help genes to express or silence.
8. ______________ is the hardware and the __________________________ is the software.
9. _________________________________ twins were used for study of epigenetics: same DNA, any
differences are due to expression.
10. Nurtured rats handled stress better: maternal behavior changed gene expression of rat offspring. Drugs can change the epigenome and _______________________________ stressed rats.
11. Human children show the same interaction. It appears that ________________________________
childhoods can promote healthy lifetime regardless of DNA: controversial. We do know the phenotype is determined by more than just the genotype for many traits.
12. _______________ can overwork stem cells: epigenetic errors accumulate through the lifetime. We
can now try to influence and correct epigenetic errors.
13. _______________________ is a neural disorder that affects information processing: social actions, communication and involves restrictive and repetitive behavior. Symptoms appear before age 3.
14. The hippocampus is linked to short term memory. It is ______________________________ in
autistic children (found by comparing twins where one twin has autism and the other does not). Perhaps methyl markers cause this…and can then be removed.
15. A study in a small Swedish village showed that when grandfathers lived in _____________________
in their late childhood, their grandsons lived longer and healthier (diabetes was studied) than those that had grandfathers that had plenty of food.
16. What was the relationship between grandmothers and granddaughters?
17. A study was done with pesticide exposure; what did it find?
18. Name one major US university that is now working to map the epigenome:
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Genetics Lab ~ Faces
Part A. Human Genetics Traits
In this exercise, you will have the opportunity to see what kind of offspring you could produce with your lab partner. Since this is strictly a pen and paper lab, same sex couples are OK. For the sake of determining the sex of your “child” one partner must be “mom” and one partner must be “dad”. Instructions
1. For the following traits, choose the phenotypes which best describes you and record the appropriate genotype for each trait in the first column on the chart at the end of the packet. If more than one genotype is possible for a given trait, choose the heterozygous condition.
2. Record your partners genotype in the second column 3. Choose which one of you tow will be the “father” and that person gets to flip for to determine
the sex of the child HEADS (X) = Female
TAILS (Y) = Male 4. For each of the remaining traits, you must donate one allele to your offspring and your partner
will donate one allele. a. If you are homozygous for a trait, you have no choice as to what allele to donate. If you
are heterozygous for a trait you must flip a coin to determine whether your child receives the dominant or recessive allele.
HEADS = DOMINANT TAILS = RECESSIVE
5. Record the genotype of your child for each trait 6. Determine your child’s phenotype for each trait 7. Using the results of your “genetic crosses” sketch the face of your child.
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Phenotypes and Genotypes
Sex Female XX Male XY Face Shape Round RR or Rr Squarish rr Chin Shape Very prominent PP or Pp Less prominent pp Dimple on Chin Present DD or Dd Absent dd Earlobes Free EE or Ee Attached ee Widows peak Present WW or Ww Absent ww Hair Texture **incomplete dominance** Curly CC Wavy Cc Straight cc Hair Color **Polygenic inheritance** Black AABB or AABb Brown AaBB or AaBb Dark (brownish) blonde aaBB Regular blonde Aabb or aaBb Pale Yelow blonde aabb Red AAbb
Eyebrows I Bushy BB or Bb Fine bb Eyebrows II (color) **incomplete dominance** Darker than Hair HH Same Color as Hair Hh Lighter than Hair hh Eyebrows III Not connected NN or Nn Connected (unibrow) nn Eye I (Distance) **incomplete dominance** Close together DD Average distance Dd Far apart dd Eye II (size) **incomplete dominance** Large EE Medium Ee Small ee Eye III (shape) Almond (wide) AA or Aa Round (narrow) aa Eye Color **Polygenic inheritance** Intense Brown AABB or AABb Brown AaBb or AAbb Hazel AaBB Green aaBB Gray blue Aabb Dark Blue aaBb Pale blue aabb
Eyelashes Long LL or Ll Short ll Lips Thick LL or Ll Thin ll Tongue Rolling Present RR or Rr Absent rr Dimples Present DD or Dd Absent dd Nose **incomplete dominance** Large NN Medium Nn Small nn PTC tasting Ability Present TT or Tt Absent tt Pinky Finger Bent BB or Bb Absent bb Thumb Straight SS or Ss Hitch Hikers ss
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Genetic Traits Overview
Widow’s Peak: Some people exhibit a hair line that drops down and forms a distinct point in the center of the forehead. This is the result of the dominant allele (W).
Tongue Rolling: A dominant allele ® gives some people the ability to roll their tongue in to a U-shape when it is extended from the mount.
Bent Little Finger: A dominant allele (B) causes the last joint of the little finger to bend inward toward the fourth finger. Lay both hands on the table, relax the muscles and note whether you have a straight or bent little finger.
Straight thumb: straight thumbs are due to a dominant allele (S). However, some people have what is commonly known as “Hitch-hiker’s Thumb” and more correctly known as distal hyperextensibility. This can be determined by bending the thumb back as far as possible. Some people can bend it back to almost a 45 degree angle. This ability is due to a recessive allele (s). There is some variation in the expression and sometimes this trait is only found in one thumb.
P.T.C. Tasting: The ability to taste P.T.C. (phyenlthiocarbamide) is associated with the function of the thyroid gland. Some people detect a distinct bitter taste from the chemical. Others do not taste it at all. Place a piece of P.T.C. paper on your tongue. If you do not detect a taste, chew the paper for a minute. If you still do not taste the paper, you are a non-taster (tt). If you are a taster, you will know it by sensing a very strong, possibly bitter taste. The ability to taste appears to be due to two allelic dominant genes, one for early tasting (Te) and one for late tasting (Tl).
Ear lobes: In most people, the ear lobes hang free. This trait is due to the dominant allele (E). However, when a person is homozygous for the recessive allele (e), the earlobes are attached to the side of the head.
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Use the Phenotypes and Genotypes page to fill out the following table:
Trait
Your Partners Child Child
Genotype Genotype Genotype Phenotype
Sex
Face Shape
Chin Shape
Dimple in Chin
Earlobes
Widow’s Peak
Hair Texture
Hair Color
Eyebrows I
Eyebrow II
Eyebrow III
Eyes I
Eyes II
Eyes III
Eye Color
Eye Lashes
Lips
Tongue Rolling
Dimples
Nose
PTC Tasting
Pinky Finger
Hitch Hikers Thumb Now that you know what your child will look like, draw a picture to represent all the traits noted above. Don’t forget to list your names as the parents! Use an 8.5”x11” piece of paper for your drawing; it should be in color!
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Additional Epistasis Problem:
In certain plants, petal color is determined by a combination of genes, where:
dd- light red + ww = petal pigment dd- light red + Ww or Ww = no petal pigment
Dd (dark red) or DD (dark red) + ww = petal pigment Dd (dark red) or DD (dark red) + Ww or Ww = no petal pigment
Hint: D/d – petal color W/w – petal pigment possible
1. Can two non-pigmented petal plants produce a light red plant?
_____________________________ (parental cross)
2. Can two dark red pigmented plants produce a plant with no pigment?
_____________________________ (parental cross)
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Karyotyping With Magnetic Chromosomes
Some background information: • Human beings have 46 chromosomes
o 22 pairs of autosomes o 1 pair of sex chromosomes
• One member of each human chromosome pair is inherited from the mother (maternal) and the other from the father (paternal)
• The maternal-paternal pair are referred to as homolouges, or homologous chromosomes • Normal human gametes carry one member of each homologous chromosome pair.
o Sometimes errors occur that create gametes with an improper number of chromosomes (numerical abnormality) or an abnormal chromosome structure (structural abnormality).
o Usually when these gametes participate in fertilization they produce nonviable embryos or offspring with disorders
• Most chromosomal abnormalities result in errors during mitosis • Numerical chromosomal abnormalities result from an incorrect distribution of chromosome into the
gametes during meiosis • This can happen when the homologous chromosomes fail to separate during meiosis I or when
sister chromatids fail to separate during meiosis II.
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Chromosomal Disorders: • When gametes with extra chromosomes are joined with a normal gametes, they produce zygotes
with three copies of that chromosome called Trisomy (trisomy 21). o 3 copies of chromosome 21 causes Down Syndrome
Most other trisomies on autosomes result in death of the embryo o A male with an extra X has Kleinfelters syndrome (XXY)
• As a rule a body can tolerate excess genetic material better than less. • Deletions of entire chromosomes usually results in a miscarriage
o Monosomy of the X chromosome is not lethal o It is called Turner’s Syndrome (XO)
Karyotypes: • Cytogenics is the study of chromosomal abnormalities and related diseases. • A Karyotype is a picture of all of the chromosomes from a somatic cell.
o A karyotype is made by obtaining one cell and using colchicines to stop cell division in metaphase of mitosis
o A dye is added to stain the chromosomes in banding patterns o A camera takes a picture under the microscope o This is called a metaphase spread
• Non-homologous chromosomes differ in size, centromere position, chromosome arm length and
staining pattern. • Homologous chromosomes have the same size, centromere position, chromosome arm length and
staining pattern. • Chromosomes are numbered from largest to smallest (1-22) and are aligned with the short arm on
top. • The sex chromosomes are traditionally located in the lower, right hand corner of the karyotype.
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Directions: You are a cyogeneticist, and just received a metaphase spread from an expecting couple. They are currently expecting, and are worried about chromosomal abnormalities. You will receive a metaphase spread of magnetic chromosomes and a magnetic layout board.
1. Remove the magnetic chromosomes form the container and spread them out face up on your lab bench. This is your metaphase spread.
2. Choose 1 magnet. Compare it to the chromosome key and identify which chromosome it represents
using your chromosome key as a guide. Look carefully at all distinguishing characteristics such as chromosome size, centromere position, length of chromosome arm, and banding pattern.
3. Once you identified the chromosome, place it on the magnetic layout board, just above the short,
solid black line with the chromosome number (or letter for sex chromosomes) below it. Line up the centromere with the long dashed horizontal line
4. Search the chromosome spread to find the homologous pair for the chromosome you identified.
Place this magnetic chromosome next to its homolog on the board.
5. Continue until you have identified all of the chromosomes.
6. In the space labeled genetic makeup below, write down the total number of chromosomes the individual has. Next indicate whether the individual represented is male or female and whether the individual has a normal or abnormal karyotype. If the karyotype indicates a chromosomal abnormality, identify the disorder possessed by the individual.
Genetic Makeup (# chromosomes)
Gender Normal/Abnormal Karyotype
Type of Chromosomal Disorder (if any)
7. Show your karyotype to the class and explain your findings.
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Questions:
1. What outcome do you expect for the individual you karyotyped? What would you tell their parents about their life expectancy and quality of life? Would you suggest the parents continue with the pregnancy or abort the pregnancy?
2. Refer to the completed karyotype display that you assembled. How is the centromere positioned on Chromosome 1, Chromosome 9, and Chromosome 14?
3. If a human gamete with an extra chromosome participates in fertilization with a normal human gamete what condition results? How many chromosomes will the zygote have?
4. If a human gamete that is missing a chromosome participates in fertilization with a normal human gamete, what conditions result? How many chromosomes will the zygote have?
5. Many genetic disorders are caused by errors (mutations) in the genes on chromosomes. Usually in disorders caused by chromosomal abnormalities, the genes themselves are not mutated, rather there are too many or two few of them. Why do you think having too many or two few normal genes creates disorders?
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Chromosome Key
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Figure 1 Figure 2
Figure 3 Figure 4
Real-World Biology Lab:
Kernel Color in Corn Corn is a good organism for studying genetics because many phenotypes can be seen in an ear of corn. Also, corn plants are easy to work with, and crosses can easily be made. Because each kernel is a separate seed, a single ear of corn contains many offspring. The more offspring you can count from a cross, the closer your experimental results will beto the theoretical results that a Punnett square predicts. In this activity, you will first work with Punnett squares to find the predicted ratios of red and yellow kernels resulting from different types of crosses, including a monohybrid cross. Then you will examine an ear of corn that resulted from a monohybrid cross to find the actual ratio of red and yellow kernels. Part A: Using Punnett Squares Red kernel color results from a dominant allele, R. The homozygous dominant kernel, RR, and the heterozygous kernel, Rr, are both red. The homozygous recessive kernel, rr, is yellow. You will use Punnett squares to predict the theoretical results of various crosses. Procedure
1. Fill in the Punnett square in Figure 1 to show a cross between a homozygous dominant parent and a homozygous recessive parent. Use the letters R and r to represent the alleles.
2. Fill in the Punnett square in Figure 2 to show a cross between a homozygous dominant parent and a heterozygous parent. Use the letters R and r to represent the alleles.
3. Fill in the Punnett square in Figure 3 to show a cross between a homozygous recessive parent and a heterozygous parent. Use the letters R and r to represent the alleles.
4. Fill in the Punnett square in Figure 4 to show a cross between two heterozygous parents, called a monohybrid cross. Again, use the letters R and r to represent the alleles.
Analyze and Conclude. Respond to each question.
5. Identify What are the possible genotypes and phenotypes of the kernels resulting from the cross shown in Figure 1? _________________________________________________________________________________
6. Calculate What are the possible genotypes, phenotypes, and predicted genotypic and phenotypic ratios of the kernels resulting from the cross shown in Figure 2? In Figure 3? In Figure 4? ___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
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Part B: Looking at Corn
Now that you have determined the predicted phenotypic and genotypic ratios, you will examine an ear of corn resulting from a monohybrid (heterozygous × heterozygous) cross and compare the observed phenotypic ratio to the predicted ratio. Procedure
1. Read and complete the lab safety form. 2. Obtain an ear of corn that is the result of a
monohybrid cross. 3. Count the number of red and yellow kernels
on the ear, and record the numbers in Table 1. Record the total number of kernels counted.
Analyze and Conclude. Respond to each question and statement.
1. Calculate the ratio of red to yellow kernels on the ear of corn. To find the ratio, divide the number of red kernels by the number of yellow kernels and round off to the nearest whole number. This number, when compared to one, is the ratio.
2. Compare your observed ratio with the theoretical ratio you predicted from the monohybrid cross shown in Figure 4. _________________________________________________________________________________
3. Hypothesize Would you have calculated the same ratio if you had counted only half the kernels on the ear of corn? Explain. _________________________________________________________________________________
4. Apply How could you determine whether a particular red kernel is homozygous dominant or heterozygous? Hint: Look at the Punnett squares in Part A. _________________________________________________________________________________
• 1-2% of the human population o 20.5% in Northern or Western Europe
• The term redhead (or red hede) has been around at least since 1510 • Scientists are researching a linkage between red hair and skin cancers such as
melanoma, basal cell carcinoma and squamous cell carcinoma Red Hair Genetics
• People with red hair have two copies of a recessive gene on chromosome #16 o Causes a mutation in the MC1R protein o This causes 80% of red hair, the other 20% comes from other genes
Usually on chromosome #4, the HCL2 gene • The MC1R protein is also linked to eye and skin color • Being a redhead is also associated with fair skin, lighter eye colors, freckles and UV sensitivity • The recessive gene codes for pheomelanin – a red pigment • The “functioning” version is called eumelanin – it is a dark pigment • Black blonde is a different amount of eumelanin • Scotland – 13% redhead; 40% carriers • Ireland – 10% redhead, 46% carriers • Heterozygote for the red hair gene have an increased sensitivity to the sun • The red hair gene increases the ability to freckle and decreases the ability to tan
Evolution
• Red hair allele mutated between 20,000-100,000 years ago • Blonde hair allele mutated between 10,000-11,000 years ago
Stereotypes
• Polynesians – red hair is a sign of decent from a high ranking ancestor or a mark of rulership • Many people Ashkanazi Jewish decent have red hair
o Possibly from an influx of European DNA o If you are reading Shakespeare or Dickens and they talk about a red headed person they are
referring to a Jewish person o During the Spanish Inquisition, red hair was considered a sign you were Jewish
Red Hair and Pain Tolerance
• There may be a link between people with red hair and tolerance for pain • Red haired people may be more susceptible to thermal pain, but less susceptible to noxious
stimulus (pinching, electrical simulation, chemical irritants) • The MC1R gene also codes for a receptor protein
